这是一份manchester曼切斯特大学 MATH20132作业代写的成功案例
$$
g(x)=\sqrt{2 x+4}, \quad x \in[0,6] .
$$
The domain of $g$ is given as the closed interval $[0,6]$. At $x=0, g$ takes on the value 2 :
$$
g(0)=\sqrt{2 \cdot 0+4}=\sqrt{4}=2 \text {; }
$$
at $x=6, g$ has the value 4 :
$$
g(6)=\sqrt{2 \cdot 6+4}=\sqrt{16}=4
$$
The most elementary way to sketch the graph of a function is to plot points. We plot enough points so that we can “see” what the graph may look like and then connect the points with a “curve.” Of course, if we can identify the curve in advance (for example, if we know that the graph is a straight line, a parabola, or some other familiar curve), then it is much easier to draw the graph.
The graph of the squaring function
$$
f(x)=x^{2}, \quad x \in(-\infty, \infty)
$$
The points that we plotted are indicated in the table and marked on the graph. The graph of the function
$$
g(x)=\sqrt{2 x+4}, \quad x \in[0,6]
$$
MATH20132 COURSE NOTES ：
For even integers $n,(-x)^{n}=x^{n}$; for odd integers $n,(-x)^{n}=-x^{n}$. These simple observations prompt the following definitions:
A function $f$ is said to be even if
$$
f(-x)=f(x) \quad \text { for all } \quad x \in \operatorname{dom}(f) ;
$$
a function $f$ is said to be odd if
$$
f(-x)=-f(x) \quad \text { for all } \quad x \in \operatorname{dom}(f) .
$$
The graph of an even function is symmetric about the $y$-axis, and the graph of an odd function is symme tric about the origin. (Figures $1.5 .7$ and 1.5.8.)
The absolute value function is even:
$$
f(-x)=|-x|=|x|=f(x) .
$$